| Exam Board | Edexcel |
|---|---|
| Module | M2 (Mechanics 2) |
| Year | 2002 |
| Session | June |
| Marks | 12 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Projectiles |
| Type | Deriving trajectory equation |
| Difficulty | Standard +0.3 Part (a) is a standard derivation of the trajectory equation using SUVAT equations and trigonometric identities, commonly found in M2 textbooks. Parts (b) and (c) involve straightforward substitution into the derived formula and solving a quadratic equation. The question requires multiple steps but uses routine techniques with no novel problem-solving insight required. |
| Spec | 1.05j Trigonometric identities: tan=sin/cos and sin^2+cos^2=13.02d Constant acceleration: SUVAT formulae3.02i Projectile motion: constant acceleration model |
| Answer | Marks | Guidance |
|---|---|---|
| \(x = u \cos \alpha t\); \(y = u \sin \alpha t - \frac{1}{2}gt^2\) | ||
| Eliminating \(t\): \(y = u \sin \alpha \frac{x}{u \cos \alpha} - \frac{1}{2}g\frac{x^2}{(u \cos \alpha)^2}\) | ||
| \(y = x \tan \alpha - \frac{gx^2}{2u^2 \cos^2 \theta}\) | ||
| \(y = x \tan \alpha - \frac{gx^2}{2u^2}(1 + \tan^2 \alpha)\) * |
| Answer | Marks | Guidance |
|---|---|---|
| \(-2 = x \tan 45° - \frac{9.8 \times x^2}{2 \times 14^2}(1 + \tan^2 45°)\) | ||
| Simplifying "correctly" to quadratic of form \(ax^2 + bx + c = 0\) (may be implied, e.g. \(x^2 - 20x - 40 = 0\); \(-0.05x^2 + x + 2 = 0\); \(4.9x^2 - 98x - 196 = 0\)) | ||
| Solving for \(t\) (2.205 s), \(x = 14 \cos 45° t\), \(x = 21.8\) m |
| Answer | Marks | Guidance |
|---|---|---|
| \(21.8_c = 14 \cos 45° t\); \(t = 2.2\) s |
## Part (a)
| $x = u \cos \alpha t$; $y = u \sin \alpha t - \frac{1}{2}gt^2$ | | | | B1; B1 |
| Eliminating $t$: $y = u \sin \alpha \frac{x}{u \cos \alpha} - \frac{1}{2}g\frac{x^2}{(u \cos \alpha)^2}$ | | | | M1 |
| $y = x \tan \alpha - \frac{gx^2}{2u^2 \cos^2 \theta}$ | | | | M1 |
| $y = x \tan \alpha - \frac{gx^2}{2u^2}(1 + \tan^2 \alpha)$ * | | | | A1 (5) |
## Part (b)
| $-2 = x \tan 45° - \frac{9.8 \times x^2}{2 \times 14^2}(1 + \tan^2 45°)$ | | | | M1 A1 |
| Simplifying "correctly" to quadratic of form $ax^2 + bx + c = 0$ (may be implied, e.g. $x^2 - 20x - 40 = 0$; $-0.05x^2 + x + 2 = 0$; $4.9x^2 - 98x - 196 = 0$) | | | | M1 |
| Solving for $t$ (2.205 s), $x = 14 \cos 45° t$, $x = 21.8$ m | | | | M1 A1 (5) |
## Part (c)
| $21.8_c = 14 \cos 45° t$; $t = 2.2$ s | | | | M1 A1 (cao) (2) |
**Total: 12 marks**
---A particle is projected from a point with speed $u$ at an angle of elevation $\alpha$ above the horizontal and moves freely under gravity. When it has moved a horizontal distance $x$, its height above the point of projection is $y$.
\begin{enumerate}[label=(\alph*)]
\item Show that
$$y = x \tan \alpha - \frac{gx^2}{2u^2}(1 + \tan^2 \alpha).$$
[5]
\end{enumerate}
A shot-putter puts a shot from a point $A$ at a height of 2 m above horizontal ground. The shot is projected at an angle of elevation of 45° with a speed of 14 m s$^{-1}$. By modelling the shot as a particle moving freely under gravity,
\begin{enumerate}[label=(\alph*)]
\setcounter{enumi}{1}
\item find, to 3 significant figures, the horizontal distance of the shot from $A$ when the shot hits the ground, [5]
\item find, to 2 significant figures, the time taken by the shot in moving from $A$ to reach the ground. [2]
\end{enumerate}
\hfill \mbox{\textit{Edexcel M2 2002 Q5 [12]}}